Continuous Rings with Chain Conditions*

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JOURNAL OF PURE AND APPLIED ALGEBRA

ELSEVIER Journal of Pure and Applied Algebra 97 (1994) 325 ~332

Continuous rings with chain conditions*

W.K. Nicholson”,“, M.F. Yousifb

Communicated by C.A. Weibel: received 31 March 1993

Abstract

The quasi-Frobenius rings are characterized as the left continuous rings satisfying either (A,) or (Al) and either (S,) or (S,), where these conditions are defined as follows: (A,): ACC on left annihilators; (AZ): R/&X(,$) is left Goldie; (S,): S = r(/(S)) for every minimal right ideal S; and (S,): Every minimal right ideal is essential in a summand of RR. These characterizations extend several results in the literature. In addition, it is shown that, in these rings, Soc(RR) = Soc( RR), Soc(eR) is simple for every primitive idempotent e of R. and there exists a complete set of distinct representatives [Rt,, , Rtn} of the isomorphism classes of the simple left R-modules such that (t, R, , t,R) is a complete set of distinct representatives of the isomorphism classes of the simple right R-modules.

1. Introduction

A ring R is called left continuous if every left ideal of R is essential in a direct summand of R and every left ideal of R isomorphic to a direct summand of R is itself a direct summand of R. Continuous rings were introduced by Utumi in [lo] as generalizations of self-injective rings. In [l 11, it was shown by Utumi that a two-sided continuous two-sided Artinian ring is self-injective (and hence quasi-Frobenius). Utumi’s results have been extended by several authors. It is known that if R is a ring such that R has ACC on left annihilators or R/Soc(,R) is left Goldie and either (a) R is left self-injective or (b) R is left and right continuous, then R is quasi-Frobenius (see [l, 2, 4, 61). IVote that there are examples of one-sided continuous two-sided Artinian rings which are not quasi-Frobenius [6]. If R is left self-injective then R is left continuous and A = r(l(A)) for every finitely generated right ideal A. If R is right continuous then, in particular, for every right ideal B of R there exists an idempotent

“Supported by NSERC Grant A8075. *Corresponding author.

0022-4049/94;$07.00 (: Elsevier Science B.V. All rights reserved SSDI 0022-4049(94)00017-D 326 W.K. Nicholson. M.F. Yous~~/Journul o/‘Pure and Applied AI&m 97 (1994) 325-332 e in R such that B is essential in the right R-module eR. All the above results can be generalized as follows.

Theorem 1. The following stutements ure equicalent,for u ring R: (i) R is quusi-Frobenius. (ii) R is kft continuous with ACC on left annihilators und S = r(l(S)) for etlery minimal right ideal S of R. (iii) R is kft continuous with ACC on left annihilutors und etlery minimul right ideal qf R is essential in eR,ftir some idempotent e in R. (iv) R is kft continuous, R/Soc(,R) is kft Goldie und S = r(l(S)),for ez;erJl minimal right ideul S of R. (v) R is left continuous, R/Soc(,R) is left Goldie und ecery minimal right ideul ofR is essentiul in eR,for some idempotent e in d.

Armendariz and Park [2, Theorem 41 showed that, if R is left self-injective and R/SocRR is left Goldie, then R is quasi-Frobenius. This follows immediately from (iv) because left self-injective rings are left continuous and also satisfy rl(1) = I for every finitely generated right ideal I. Moreover, (iii) generalizes a result of Camillo and Yousif [4, Theorem 1] and (v) generalizes a result of Ara and Park [I, Corollary 2.71. A key result in the proof of Theorem 1 is a theorem in which the finiteness conditions are relaxed. More precisely, we show that if R is left continuous, R/Soc(,R) has ACC on left annihilators, and S = rl(S) for every minimal right ideal S of R, then R is a semiprimary ring with Soc( RR) = Soc(R,). We also provide an example of a commutative semiprimary local continuous ring R with .uR = unn(unn(.yR)) for every x E R, but which is not self-injective. Throughout this paper all rings considered are associative with identity and all modules are unitary R-modules. We write J(M), Z(M), E(M), and Sot(M) for the Jacobson radical, the singular submodule, the injective hull and the socle of M, respectively. For any subset X of R, l(X) (resp. r(X)) represents the left (resp. right) annihilator of X in R. The notation A ‘c B will mean RA is essential in KB as left R-modules. For a full account of continuous rings and continuous modules we refer the reader to [S, 10, 111.

2. The results

Recall that a ring R is called right Kasch [7] if every simple right R-module is isomorphic to a minimal right ideal of R, equivalently if I(!) # 0 for every proper right ideal I of R. Left Kasch rings are defined analogously.

Lemma 1 (Ara and Park [l, Proposition 1.41). [fR is left continuous und SAC is ,jinitely generated und es.sential in RR, then R is u semipe~fkct kft Kusch ring and SOC(~R) c_ Soc(R,). W.K. Nicholson, M.F. Yousif/Journal of’Pure and Applied Algebra 97 (1994) 325-332 327

A submodule N of a module M is said to lie over a summand of M if M = P@Q where P c N and NnQ is small in M. The following result is well known (see [S, Corollary 4.421).

Lemma 2. If R is semiperfect, ez;ery kft or right ideal lies over a summand qf R.

Lemma 3. Suppose R is semipecfect with R(Soc(RR)) ‘e KR. Then R is a right Kasch riny.

Proof. Let T be a maximal right ideal of R. Since R is semiperfect, there exists (by Lemma 2) are idempotent e of R such that eR G T and Tn(1 - e)R is small in R. Since eR 5 T we can write T = eR @( Tn( 1 - e)R). And since Tn( 1 - e)R is small in R, Tn( 1 ~ e)R s J(R), and hence 1(J) E I( T n (1 - e)R). Since SocRR G I(J), it follows by hypothesis that I(Tn(1 - e)R) ? RR. But then I(eR) = R(l - e) # 0 implies that I(T) = I(eR@ Tn( 1 - e)R) = I(eR)nl(T n (1 - e)R) # 0, which completes the proof of Lemma 3. q

Lemma 4. Suppose R is a left continuous riny and R/Soc(,R) hus ACC on kji unnihi1utor.s. Then: (i) R is a semiprimary ring. (ii) Soc(Re) is simple ,for every primitice idempotent e of R. (iii) So&R) E Soc(RR). (iv) R is a two-sided Kasch riny. (v) rl( T, n ... n T,) = T, n ... n T,, ,fbr any finite ,family of muximul right ideals T,;.., T, of R. (vi) Ir(L,n ‘.. n L,) = L,n ... n L,, .fk uny ,$nite ,fumily of maximal kft ideuls L,;..,L, ofR. (vii) h(J) = r/(J) = rr(J) = J where J = J(R). (viii) Z(R,) G J(R) = Z(,R). (ix) Z(R,) = I(SocRR) and Z(RR) = r(Soc,R).

Proof. (i) By [I, Theorem 1.21. (ii) Since R is semiprimary, SocRR is essential in RR. Now if e is a primitive idempotent and S is a simple submodule of Re then S is essential in a summand of Re. Since Re is indecomposable, Soc(Re) is simple. (iii) This follows from (i), (ii) and Lemma 1. (iv) By Lemma 1, R is a left Kasch ring. That R is right Kasch follows from (i), (iii), and Lemma 3. (v) Since R is right Kasch, T = r/(T) for every maximal right ideal T of R. Now if T1, . . . , T, is any family of maximal right ideals of R, then /(T,) + ... + /(T,) z I(T,n ..t nT,J. Thus rl(T,n ... n T,) G r(l(T,) + ‘.. + I(T,)) = r/(T,)n ... n rl(T,) = T1 n ... n T,,. Since we always have T,n ... nT, 5 r/(T, I-I ... nT,), the desired result follows. 328 W.K. Nicholson, M.F. Yousif/Journal q/Pure and Applied Algebra 97 (1994) 325-332

(vi) The proof is similar to the one given in (v) above. (vii) Since J is an intersection of a finite family of maximal left (resp. right) ideals we infer, from (v) and (vi) above, that r/(J) = /r(J) = J. Let .K E U(J). Then r(J). x = 0 and hence (Soc,R).u = 0. Since SocRR ‘e RR, .Y E .Z(,R). Since R is left continuous, Z(,R) = J and so .Y E J. Hence U(J) 5 J; the other inclusion holds since J E Z(,R) G r(So&R)) because R is left continuous, and Soc(RR) = r(J) because R is semiprimary. (viii) It is easy to see that if R is any semiperfect ring then Z(R,) G J. For, in this case Z(R,) lies over a summand of R by Lemma 2. That is, there exists an idempotent e of R such that eR G Z(R,) and (1 - e)Rn Z(R,) is small in R. Since Z(R,) does not contain any non-zero idempotents, it follows that Z(R,) 5 J. We have J = Z(,R) because R is left continuous. (ix) This is clear since SocR, ‘e RR and Soc,R ‘2 RR by (i). 0

Theorem 2. Suppose R is left continuous ring suck tkut R/Soc,R kus ACC on bft annihilators. If xR = rl(xR) .for every minimal right ideal xR of R, then Soc( RR) = Soc(R,), Soc(eR) is simple for every primitioe idempotent e of R und there exists a complete set qf distinct representaiues {R t, , , Rt,) of the isomorpkism classes qf the simple lgft R-modules suck tkut it, R, , t, R) is u complete set qf distinct representutices of the isomorpkism classes of the simple right R-modules.

Proof. Since R is semiprimary, SocRR ‘e RR and SocRR ‘2 RR, and R = Re, @ ... @ Re, where {Re,, . . . , Re,}, n I m, is a complete set of distinct representatives of the indecomposable projective left R-modules. Thus {e, , . , e, ) is a basic set of primitive idempotents for R. Write Tk = Soc(Rek), 1 I k i n; Tk is simple by Lemma 4(ii). We claim that ek. Tk # 0 for some k’, 1 i k’ I n. For, if eiT, = 0 for every i, 1 I i I n, then eT, = 0 where e = ei + ... + e, is the basic idempotent; whence RT,, = (ReR)T, = 0, a contradiction. So choose k’, 1 I k’ I n, with ek,T, # 0, and let 0 # tk E ek’ Tk. Clearly l(t,R) 2 J + R(l - ek.), where J = J(R). Since J + R(l - Q.) is a maximal left ideal of R, l(t,R) = J + R(l - ekZ). Now, let hR be any minimal right ideal of ek. R. We claim that l(bR) is a maximal left ideal of R. For, if L is a maximal left ideal of R containing l(hR), then r(L) # 0 since R is left Kasch ring. Thus r(L) G rl(hR) = hR implies that r(L) = hR, from which it follows (using Lemma 4) that L = Iv(L) = I(hR), proving the claim. Hence, since R( 1 - ekZ) G l(bR) and J + R( 1 - ek.) is the unique maximal left ideal of R containing R(l - ek’), it follows that l(hR) = J + R(l - ek.). But then, 0 # t,R c rl(t,R) = r(J + R(l - +)) = rl(bR) = hR. Thus t,R is the unique minimal right ideal of ek3R. We now claim that t,R 2 e,R/e,J. For if a:ekR + t,R is defined by X(X) = tk.x, then x is a non-zero map since c((ek) = tk, and c( is onto because t,R is simple. Since ker(a) = ekJ, our claim follows. Now the map from ( 1, 2, , n) to itself which sends k to k’ is actually a (Nakayama) permutation of ( 1, , n). To see that it is one-to-one, 1et.j’ = k’. Then t,R = Soc(e,, R) = Soc(ej, R) = t,jR and, if r + rdenotes the canonical W.K. Nicholson. M.F. Yous~f/Journal ofPurr and Applied Algebra 97 (1994) 325-332 329 quotient map R + R/J, then gkR z t,R g eja and hence j = k. Thus Soc(e,R) is simple for every k, 1 I k I n. Now recall that Rt, = Soc(Rek), 1 < k I n. We claim that {Rt,, , Rt,) is a set of representatives of the isomorphism classes of the simple left R-modules. For if Rt, z Rtj, 1 5 k, j 5 n, then E(Re,) = E(Rt,) % E(Rtj) = E(Re,) and by 17, Corollary 2.141 it follows that Rek r Rej and hence that k = j. Since every simple left R-module is embedded in R (Lemma 4(iv)), our claim follows. Now, from the above work, since each tk R z ekR/ekJ, it follows similarly that (tl R, , t, R) is a set of representatives of the isomorphism classes of the simple right R-modules. Finally we want to show that SocRR = SocRR. From Lemma 4(iii), it is enough to show that SocR, c SocRR. Let Soc(R,) = CL 1xiR, where each xiR is a minimal right ideal of R. Since each XiR = r(&) for some maximal left ideal Li of R, Soc(R,) = X7= 1 r(Li). Thus I(SocRR) = l(Cy=, r(&)) = fly= 1 lr(,!+) = fly=, Li, and hence J E n$ 1 Li = l(SocRR) = Z(R,), by Lemma 4(ix). Now, Lemma 4(viii) gives J = Z(RR). Since J. Soc(R,) = Z(R,).Soc(R,) = 0, we infer that SocRR L r(J) = SocRR, which completes the proof of Theorem 2. 0

For convenience, a module MR is called a min-CS module if every simple submodule of M is essential is a summand of M. A ring R is called a right min-CS ring if RR is a min-CS module. Clearly every CS-module is a min-CS module. However the converse is not true, see [9, Example 1.11.

Lemma 5. Suppose R is l$’ continuous, RISoc,R has ACC on left annihilators and R is u right min-CS ring. Then xR = rl(xR) ,for every minimal right ideul xR of R. In purticulur SocRR = SocRR.

Proof. Let xR be a minimal right ideal of R. Then .uR ‘e eR, for some e2 = e E R. Since xR E rl(xR) G rl(eR) = eR, we infer that xR ‘c rl(xR). Let L be a maximal left ideal of R containing I(xR). By Lemma 4, it follows that 0 #r(L) E r(J) = SOC(~R) G Soc(RR), and hence that r(L) is a non-zero semisimple right ideal of R. Moreover 0 # r(L) c r/(xR) and xR E rl(xR), so xRnr(L) # 0. Thus xR 2 r(L), and so xR = r(L), from which it follows that l(xR) = lr(L) = L and xR = rl(xR). Now from Theorem 2, it follows that SocRR = SocRR. 0

Proof of Theorem 1. (i) 3 (ii), (iii), (iv), and (v) is clear. (ii) 3 (i) Since R has ACC on left annihilators, Z(,R) = J(R) = J is nilpotent, and R has no infinite set of orthogonal idempotents. Since R is left continuous, R/J is regular and hence semisimple Artinian. Therefore R is a semiprimary ring. Now, since SocRR = r(J) and R has DCC on right annihilators, it follows that R/Soc,R has DCC on right annihilators and hence ACC on left annihilators. From Theorem 2, it follows that SocRR = SocR, and SocR is left and right finite dimensional. Now an application of Lemma 6 of [4] will ensure that R is a two-sided Artinian ring. Since Soc(eR) and 330 W.K. Nicholson. M.F. Yousif/Journal of Purr and Applied Algebra 97 (1994) 325-332

Soc(Re) are simple for every primitive idempotent e of R, it follows that R is quasi-Frobenius [7, p. 3421. (iii) * (ii) As in the proof of (ii) * (i), R/Soc(R,) has ACC on left annihilators. Now (i) follows from Lemma 5. (iv) * (i) From Lemma 4(i) R is semiprimary, and from Theorem 2 SocRR = SocRR and hence R is left and right finite dimensional. Since R/&CR is left Goldie, Soc(R/SocR) = Soc2(R)/Soc(R) is left Artinian and hence Sot,(R) is left Artinian. By [l, Theorem 2.21, R is two-sided Artinian, and hence R is quasi- Frobenius 17, p. 3421. (v) * (i) This follows from Lemma 5. 0

It is also known 12, Theorem 41 that if R is left self-injective and R/Soc(R,) is left Goldie then R is quasi-Frobenius. With the help of Theorem 2 and the arguments used in [2, Theorem 41 we will also extend this result.

Proposition 1. Suppose R is left continuous und R/Soc(R,) is kft Goldie. Then R is semiprimary.

Proof. A straightforward adaptation of the argument used in the proof of [2, The- orem 4(b)] will yield the result. The only place where the reader may need help is on p. 29 of the article (paragraph 3). In order to show that Soc(RR)e = A is left finite dimensional the reader may appeal to [S, Corollary 31. 0

Corollary 1. Suppose R is ltlfi continuous, R/Soc(R,) is left Goldie mnd xR = rl(xR),for every minimul right ideal .uR qf R. Then R is quasi-Frohenius.

Proof. By Proposition 1, R is semiprimary. Therefore Soc,R ‘E RR and Soc(RR) “2 RR. As before So&R) E Soc(RR). Now by Lemma 3, R is a right Kasch ring. And by Lemma 1, R is a left Kasch ring. Now with an argument similar to the one used in the proof of Theorem 2 we can show that SocRR = SocR,. By Theorem l(iv), R is quasi-Frobenius. q

Corollary 2. Suppose R is left continuous, right min-CS ring and R/Soc(R,) is i

Proof. By Proposition 1, R is semiprimary. As before R is a two-sided Kasch ring. And with an argument similar to the one used in the proof of Lemma 5, we can show that xR = rl(xR) for every minimal right ideal .uR of R. Now by Corollary 1, R is quasi-Frobenius. 0

Example. (Camillo [3, Remark 2, p. 361). Consider the ring R of polynomials in countably many indeterminates {.xi) over Zz = Z/22, where we impose the W.K. Nicholson, M.F. Yous~f/Journal of’Pure and Applied Algebra 97 (1994) 325-332 331 following relations: (i) .x2 = 0 for all k, (ii) SkXj = 0, k # j and (iii) xi = of for all k,j. As in [3], R is commutative, semiprimary, local, has simple socle and satisfies xR = ann(ann(xR)) for all x E R. But R is not self-injective. We claim that R is continuous. Since R is uniform, every ideal of R is essential in R. That is, R satisfies the first condition of continuity (CS-condition). To see that R satisfies the second condition of continuity, suppose A is an ideal of R isomorphic to a summand of R(that is A z R). We want to show A is a summand of R (i.e. A = R). Let .K E A and consider the following diagram:

inc. Q-----+R.x~ R inc.

Then, by 131, there exists a mapf: R + A such that ,f(x) = X. Let ,f‘(l) = u E A. Then s = xa E xA. Thus R/A is flat. Since R is semiprimary, R/A is projective. Hence A is a summand of R(i.e. A = R).

Acknowledgements

This research was carried out during a visit by the second author to The University of Calgary and was supported by NSERC Grant A8075. The second author would like to thank the University of Calgary for its warm hospitality. The second author was also supported by the Ohio State University in the form of a Special Research Assignment Quarter and a research grant. Both authors would like to thank the referee for several suggestions which improved the exposition of the paper.

References

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